Gravitation

JEE Main level, numerical answer, medium

30 of 223

Question 30 of 223

Acceleration due to gravity on the surface of earth is ' gg '. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ……\ldots\ldots g.

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Answer: 9

∵\because acceleration due to gravity on surface is given by
g=GMRe2\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^2}

Now since diameter is reduced to 1/3rd 1 / 3^{\text {rd }}, radius also reduces to 1/3rd 1 / 3^{\text {rd }}, keeping mass constant
New value of acceleration due to gravity on Earth's surface is
g′=GM(Re3)2=9GMeRe2=9 g\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\left(\frac{\mathrm{R}_{\mathrm{e}}}{3}\right)^2}=9 \frac{\mathrm{GMe}}{\mathrm{R}_{\mathrm{e}}^2}=9 \mathrm{~g}

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